Consider the system of equations :
$ r'(t)=3r(t)+\cos t \\ \theta'(t)=r(t)-4 \theta(t)+3 \ $
Write the above system in the matrix form $ \ x'=Ax+ f \ $
Answer:
I am confused whether $ \ x'=Ax+f \ $ means $ x=(x,y) \ \ or \ \ x=r, \theta ) \ $.
I thought here we have to express in $ \ (x,y) \ $.
Let $ \ x=r \cos \theta, \\ y=r \sin \theta, \ $
Then,
$ x'(t)=r'(t) \cos \theta-r \sin \theta \theta'(t) \ =[3r(t)+\cos t] \cos \theta-r \sin \theta [r(t)-4 \theta (t)+3] \\ y'(t)=r'(t) \sin \theta+r \cos \theta \theta'(t)=[3r(t)+\cos t] \sin \theta+r \cos \theta [r(t)-4 \theta(t)+3] $
How to eliminate $ \ \theta \ $ and $ \ r(t) \ $ from the new system in order to get the form $ \ x'=Ax+f \ $?
Help me doing this
We have that
$$x=\begin{bmatrix}r(t)\\\theta(t)\end{bmatrix}$$
thus
$$ x'=Ax+ f $$
$$\begin{bmatrix}r'(t)\\\theta'(t)\end{bmatrix}=\begin{bmatrix}3&0\\1&-4\end{bmatrix}\begin{bmatrix}r(t)\\\theta(t)\end{bmatrix}+\begin{bmatrix}\cos t\\3\end{bmatrix}$$